8.1 ${\pi }_1(S_1)$

In this section, our goal is to show that ${\pi }_1({𝕊}^1)=\mathbb{Z}$.In fact, we will show that the loop space $\mathrm{\Omega }({𝕊}^1)$ is equivalent to $\mathbb{Z}$.This is a stronger statement, because ${\pi }_1({𝕊}^1)=\parallel \mathrm{\Omega }({𝕊}^1){\parallel }_0$ bydefinition; so if $\mathrm{\Omega }({𝕊}^1)=\mathbb{Z}$, then $\parallel \mathrm{\Omega }({𝕊}^1){\parallel }_0=\parallel \mathbb{Z}{\parallel }_0$ by congruence, and$\mathbb{Z}$ is a set by definition (being a set-quotient; see defn-Z,Z-quotient-by-canonical-representatives), so $\parallel \mathbb{Z}{\parallel }_0=\mathbb{Z}$.Moreover, knowing that $\mathrm{\Omega }({𝕊}^1)$ is a set will imply that ${\pi }_n({𝕊}^1)$ is trivial for $n>1$, so we will actually have calculated all the homotopy groups of ${𝕊}^1$.